Obviously the two planes are parallel, since the normal vector of the first plane is a scalar product of the normal vector of the second plane.
Let A (0, 0, 5) be a point on the first plane.
Let B (0, 0, (71)/2) be a point on the second plane.
Then the vector AB= u = < 0, 0, (61)/2 > .
Hence the distance of the two parallel planes is equal to the length of the projection of vector AB on the
normal vector n=< -4, 5, 2 > of the planes.
Then if we denote by d the distance between the two parallel planes
we have d = ( || u⋅n || ) / ||n|| = (61)/( 3√5)
